Prove that :$[BCA']^2+[CAB']^2+[ABC']^2=[ABC]^2$, where [•] denote the area.
Hint: Prove first the following generalization of the Pythagorean theorem:
(1) If $OA,OB,OC$ are three mutually perpendicular lines in $\mathbb{R}^3$, $$ [ABC]^2=[OAB]^2+[OBC]^2+[OAC]^2 .$$
then prove that in your configuration $A'B=C'B,B'A=C'A,A'C=B'C$, hence you may conclude from the previous lemma. How to prove that $A'B=C'B$? Simple. We have: $$ A'B^2+A'C^2 = a^2, \qquad A'B^2-A'C^2 = AB^2-AC^2 = c^2-b^2 $$ (since $BB'\perp AC$) hence $A'B^2=\frac{a^2+c^2-b^2}{2}=C'B^2$.
Like Desargues' theorem, this $2d$-problem is easier to prove if we consider it in $3d$!
